 Hello and welcome to the session. In this session we discuss the following question which says Suppose vector A is equal to 2i cap minus j cap, vector B is equal to 9j cap minus 3k cap and vector C is equal to 14i cap minus 7k cap. Find vector D such that vector D is perpendicular to vector A, vector D is perpendicular to vector B and vector D dot vector C is equal to 1. Now given vectors A and B, vector A is said to be perpendicular to vector B if and only if vector A dot vector B is equal to 0. This is the key idea that we use for this question. Now we proceed with the solution. We have vector A equal to 2i cap minus j cap, vector B is equal to 9j cap minus 3k cap and we have vector C equal to 14i cap minus 7k cap. So we have to find vector D we suppose vector D be equal to D1i cap plus D2j cap plus D3k cap. Now we have given one condition that vector D is perpendicular to vector A. So vector D perpendicular to vector A means that vector D dot vector A is equal to 0. Now we have vector D dot vector A equal to 0. Now putting the values for vector D and vector A we have D1i cap plus D2j cap plus D3k cap dot vector A which is 2i cap minus j cap is equal to 0. So this means 2D1 minus D2 is equal to 0. So we have got one equation, let this be equation 1. Now the other condition given to us in the question is vector D is perpendicular to vector B, vector D perpendicular to vector B means that vector D dot vector B is equal to 0. Now vector D dot vector B is equal to 0. So we put the values for vector D and vector B this means we get D1i cap plus D2j cap plus D3k cap dot vector B which is 9j cap minus 3k cap and this is equal to 0. So further we get 9D2 minus 3D3 is equal to 0. This means we have 3D2 minus D3 is equal to 0. This is equation 2. Now the other condition given to us is vector D dot vector C is equal to 1. So we put the values for vector D and vector C. So from here we get D1i cap plus D2j cap plus D3k cap dot vector C which is 14i cap minus 7k cap is equal to 1. So from here we get 14D1 minus 7D3 is equal to 1. We take this as equation 3. So now we have got three equations. First of all we have from equation 1 2D1 minus D2 equal to 0. This means we have D2 is equal to 2D1. So now substituting D2 equal to 2D1 in equation 2 we get 3 into 2D1 minus D3 is equal to 0. This means we have 6D1 minus D3 equal to 0. So let this be equation 4. So now consider the equation 3 which is 14D1 minus 7D3 equal to 1. Now we will solve equations 3 and 4 for D1 and D3. So now multiplying equation 4 by 7 we get 42D1 minus 7D3 equal to 0. Now the other equation is 14D1 minus 7D3 equal to 1. Now solving these two equations we get 28D1 is equal to minus 1. This means D1 equal to minus 1 upon 28. We have D2 equal to 2D1. So putting the value for D1 we get D2 equal to 2 into minus 1 upon 28. Now D2 14 times is 28. So we get D2 as minus 1 upon 14. Consider the equation 4 now which is 6D1 minus D3 equal to 0. This means that D3 is equal to 6D1. Putting the value for D1 as minus 1 upon 28 here we get D3 is equal to 6 into minus 1 upon 28. Now 2 3 times is 6 and 2 14 times is 28 that we get D3 as minus 3 upon 14. So finally we get the values for D1 D2 D3. So we have D1 equal to minus 1 upon 28 D2 equal to minus 1 upon 14 D3 equal to minus 3 upon 14. Thus vector D equal to minus 1 upon 28 i cap minus 1 upon 14 j cap minus 3 upon 14 k cap. So finally we get the vector D. This is our final answer. So this completes the session. Hope you have understood the solution of this question.